Question

Show that the points $A,B,C$ with position vectors $2\widehat i - \widehat j + \widehat k,\widehat i - 3\widehat j - 5\widehat k$ and $3\widehat i - 4\widehat j - 4\widehat k$ respectively, are the vertices of a right angled triangle. Hence find the area of the triangle.

Answer 1

Hint: In this question the position vectors of the vertices are given. So, first try to find out the length of all sides of the triangle. Using Pythagoras theorem if it follows Pythagoras theorem then it will be proved that the triangle is a right angled triangle. If it is a right angled triangle then find out the area of the triangle.
Given: The points $A,B,C$ with position vectors
$\overrightarrow A = 2\widehat i - \widehat j + \widehat k$
$\overrightarrow B = \widehat i - 3\widehat j - 5\widehat k$
$\overrightarrow C = 3\widehat i - 4\widehat j - 4\widehat k$

Step-by-step solution: Since, the position vectors of all the vertex of triangles are given so first we try to find out $\overline {AB} ,\overline {BC} $ and of the given vector.
 We know that $\overline {AB} = \overline B - \overline A $
Similarly, $\overline {BC} = \overline C - \overline B $
$\overline {CA} = \overline A - \overline C $
$
  \therefore \overline {AB} = (1 - 2)\widehat i + (1 - 3)\widehat j + ( - 1 - 5)\widehat k \\
   = - \widehat i - 2\widehat j - 6\widehat k \\
  \overline {BC} = \overline C - \overline B = (3 - 1)\widehat i + ( - 4 + 3)\widehat j + ( - 4 + 5)\widehat k \\
   = 2\widehat i - \widehat j + \widehat k \\
  \overline {CA} = \overline A - \overline C = - \widehat i + 3\widehat j + 5\widehat k \\
 $
$\therefore $ Length of $\left| {\overline {AB} } \right| = \sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{( - 6)}^2}} = \sqrt {1 + 4 + 36} = \sqrt {41} $ (Formula used: $a\widehat i + b\widehat j + c\widehat k$ is given then magnitude will be $\sqrt {{a^2} + {b^2} + {c^2}} $ )
$\therefore $ Length of $\left| {\overline {BC} } \right| = \sqrt {{{(2)}^2} + {{( - 1)}^2} + {{(1)}^2}} = \sqrt {4 + 1 + 1} = \sqrt 6 $
$\therefore $ Length of \[\left| {\overline {AC} } \right| = \sqrt {{{( - 1)}^2} + {{(3)}^2} + {{(5)}^2}} = \sqrt {1 + 9 + 25} = \sqrt {35} \]
From the above obtained length it show that
$\therefore {(AB)^2} = {(BC)^2} + {(AC)^2}$
$
   \Rightarrow {(\sqrt {41} )^2} = {(6)^2} + {(35)^2} \\
   \Rightarrow 41 = 6 + 35 \\
  \therefore 41 = 41 \\
 $
Hence, the given triangle $ABC$ is a right angled triangle.
$\because $ Base = $\left| {\overline {BC} } \right|$ Height = $\left| {\overline {CA} } \right|$
Now, area of triangle $ABC = \dfrac{1}{2} \times Base \times Height$

Finding cross product
$\overline {BC} \times \overline {CA} $
\[
  \overline {BC} \times \overline {CA} = \left( {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  2&{ - 1}&1 \\
  { - 1}&3&5
\end{array}} \right) \\
   = \widehat i( - 5 - 3) - \widehat j(10 + 1) + \widehat k(6 - 1) \\
   = - 8\widehat i - 11\widehat j + 5\widehat k \\
 \]
$\left| {\overline {BC} - \overline {CA} } \right| = \sqrt {64 + 121 + 25} = \sqrt {210} $
$\therefore $ Area of triangle $ABC = \dfrac{1}{2}\sqrt {210} $ unit square
$ = \dfrac{{\sqrt {210} }}{2}$

Note: In this question vector order arrangement must be the same otherwise it will mislead the result. When we try to find out the cross product for the area then the sign should be according to proper instruction.